# F(x,y,z) = x\vec{i}+y\vec{j}+z\vec{k}; \sum is the intersection of the cylinder x^{2}+y^{2}=1...

## Question:

{eq}F(x,y,z) = x\vec{i}+y\vec{j}+z\vec{k}{/eq}; {eq}\sum{/eq} is the intersection of the cylinder {eq}x^{2}+y^{2}=1{/eq} with the plane {eq}z=y{/eq}

## Flux of a Vector Field:

If we have a vector field {eq}\mathbf F(x, y, z) = f(x, y, z) \: \mathbf{i} + g(x, y, z) \: \mathbf{j} + h(x, y, z) \: \mathbf{k}, {/eq} and a surface {eq}S {/eq} defined by the function {eq}z = f(x, y) {/eq} over the region {eq}R {/eq} in the plane, then a normal vector to {eq}S {/eq} is given by the formula {eq}\mathbf N = -f_x(x, y) \: \mathbf{i} - f_y(x, y) \: \mathbf{j} + \mathbf{k}. {/eq} The flux of {eq}\mathbf F {/eq} through {eq}S {/eq} can then be calculated by evaluating the surface integral {eq}\displaystyle\iint_S \mathbf F \cdot \mathbf N \: dS. {/eq}

We are given the vector field {eq}\mathbf F(x,y,z) = x \: \mathbf {i}+y \: \mathbf {j}+z \: \mathbf {k} {/eq}, and {eq}\sum {/eq} is the intersection of the cylinder {eq}x^{2}+y^{2}=1 {/eq} with the plane {eq}z=y {/eq} We will calculate the flux of {eq}\mathbf F {/eq} through {eq}\sum. {/eq} First, a normal vector to the plane {eq}f(x, y) = y {/eq} is {eq}\mathbf N = -f_x(x, y) \: \mathbf{i} - f_y(x, y) \: \mathbf{j} + \mathbf{k} = -\mathbf{j} + \mathbf{k}. {/eq} Next, the projection of the surface onto the {eq}xy {/eq} plane is the unit circle, centered at the origin. Therefore, the flux is

{eq}\begin{eqnarray*}\iint_\sum \mathbf F \cdot \mathbf N \: dS & = & \iint_D (x \: \mathbf {i}+y \: \mathbf {j}+z \: \mathbf {k}) \cdot (-\mathbf{i} + \mathbf{k}) \: dA \\ \\ & =& \iint_D -x + z \: dA \\ \\ & = & \iint_D -x + y \: dA \end{eqnarray*} {/eq}

Next we convert to polar coordinates, so {eq}x = r \cos \theta, \: y = r \sin \theta, {/eq} and {eq}dA = r \: dr \: d\theta. {/eq} Therefore

{eq}\begin{eqnarray*}\iint_D -x + y \: dA & = & \int_0^{2\pi} \int_0^1 (-r \cos \theta + r \sin \theta) \: r \: dr \: d\theta \\ \\ & =& \int_0^{2\pi} \int_0^1 r^2 (\sin \theta - \cos \theta) \: dr \: d\theta \\ \\ & = & \int_0^{2\pi} \displaystyle\frac{r^3}3 (\sin \theta - \cos \theta) \biggr|_0^1 \: d\theta \\ \\ & =& \displaystyle\frac13 \int_0^{2\pi} \sin \theta - \cos \theta \: d\theta \\ \\ & =& \displaystyle\frac13 \left( -\cos \theta - \sin \theta \right. \biggr|_0^{2\pi} \\ \\ & = & 0 \end{eqnarray*} {/eq} 